Relativistic Momentum from Conservation

Derivation

Relativistic Momentum from Conservation

D-098 Home PU-105 Threads matter · force · symmetry Depends on Proper Time and the Four-Velocity, Relativistic Velocity Addition

Statement

If total momentum is to be conserved in every inertial frame under the Lorentz transformation of velocities, then the momentum of a single particle of rest mass m moving with velocity v cannot be mv; it must instead be p = γmv, where γ = 1/√(1 − v²/c²). This reduces to the Newtonian p = mv when v « c.

Why it matters

Conservation of momentum is one of the deepest laws in physics, tied by Noether's theorem to the homogeneity of space. Special relativity keeps that law but changes how velocities combine between frames. The two demands are only mutually consistent for one particular momentum function, so relativity does not merely permit the factor γ — it forces it.

This is the gateway to relativistic dynamics: once p = γmv is fixed, the energy E = γmc², the invariant E² − (pc)² = (mc²)², and the whole machinery of colliders and particle physics follow.

Assumptions

Momentum is conserved in all inertial frames.If conservation held in only one frame, no constraint would couple the frames and any function p(v) would survive; the derivation collapses.

Velocities transform by the relativistic addition law (from relativistic-velocity-addition), not by Galilean addition.Under Galilean addition the ordinary mv is already frame-consistent, and no correction factor is required.

Momentum is a vector along the velocity, p = m f(v) v, with f an even scalar function of speed and f(0) = 1.Isotropy of space forbids any preferred direction other than v itself; dropping this admits terms that would make momentum depend on orientation with no physical origin.

Rest mass m is a Lorentz-invariant scalar, the same number in every frame.If m were frame-dependent it could not be factored out of the transformation, and "mass" would lose its meaning as an intrinsic property of the particle.

Derivation

We use the standard symmetric collision argument. Two identical particles, each of rest mass m, approach along the y-axis in a symmetric way and collide. We analyse the collision in two frames connected by a boost along x, and demand that the y-momentum balance in both. Write the unknown momentum as p = m f(u) u for a particle of speed u = |u|, with f to be determined.

S: two particles A, B; before collision u_A = (0, +w), u_B = (0, −w)

Set up a symmetric glancing collision in the centre-of-momentum frame S; by symmetry each particle simply reverses its y-velocity. A

S′ moves at velocity (−V, 0) relative to S, with V chosen so A moves purely along y in S′

Choose a second frame boosted along x; the boost axis is perpendicular to the collision axis, so the physics of the balance is unchanged but the velocities acquire x-components. A

transverse velocity transform: u_y′ = u_y / [γ_V(1 + u_xV/c²)]

Apply the relativistic velocity-addition law for the component perpendicular to the boost (from relativistic-velocity-addition). Here γ_V = 1/√(1 − V²/c²). B

For A: u_Ax = 0 in S, so u_Ay′ = w/γ_V, u_Ax′ = −V

Substitute u_x = 0 into the transform; the transverse speed is reduced by γ_V (time dilation of the transverse motion), and A picks up the frame's x-velocity. B

For B: u_Bx = 0 in S, so u_By′ = −w/γ_V, u_Bx′ = −V

Same transform applied to B; both particles share the common x-velocity −V in S′, while their y-velocities are equal and opposite. B

speeds in S′: u_A′² = V² + w²/γ_V² = u_B′²

Add the squared components; both particles have the same speed magnitude in S′, hence the same f(u′) — call it f′. This equality is what makes the balance tractable. A

y-momentum in S′ before: P_y′ = m f′ u_Ay′ + m f′ u_By′ = 0

The two y-velocities are equal and opposite and share the same f′, so the total y-momentum vanishes before the collision — and by the reversal symmetry, after it too. Balance in S′ is automatic. B

Now impose balance in S. Before: P_y = m f(w)(+w) + m f(w)(−w) = 0. This is guaranteed; the content is the cross-frame link.

Momentum balances trivially in each frame by symmetry. The non-trivial requirement is that the same function f describe the particle in both frames — this couples f at different speeds. A

Consistency demands f(w) w and f′ u_Ay′ describe one particle's y-momentum, transformed. Since transverse momentum is frame-invariant: m f(w) w = m f′ (w/γ_V)

The y-component of momentum is unchanged by an x-boost (transverse momentum is a Lorentz invariant). Equate A's y-momentum in S and in S′. This is the equation that pins down f. C

cancel m w: f(w) = f′ / γ_V

Divide through by the common m w (nonzero for w ≠ 0). We now have f at speed w in terms of f at the larger speed u′ and the boost factor. B

Take the limit w → 0. Then f(w) → f(0) = 1, and from step 6 u′² → V², so f′ → f(V)

In this limit the transverse motion is negligible and A moves at speed V in S′. The normalisation f(0) = 1 fixes the low-speed end (Newtonian limit). B

1 = f(V) / γ_V ⇒ f(V) = γ_V = 1/√(1 − V²/c²)

Substitute the limits from step 11 into step 10. The unknown function is forced to be exactly the Lorentz factor evaluated at the particle's speed. C

Rename V → v (any speed): p = m f(v) v = γ m v

V was an arbitrary boost speed, so the result holds for a particle of any speed v. This completes the identification of the momentum function. A

Result

p = γ m v = m v / √(1 − v²/c²)

Reading. The relativistic momentum is the Newtonian expression mv multiplied by the Lorentz factor γ ≥ 1. As v → c, γ → ∞, so momentum grows without bound even though speed is capped at c — which is why no finite impulse can push a massive particle to light speed. The factor is not an add-on; conservation in all frames leaves no other choice.

Units check. γ is dimensionless, so [p] = [m][v] = kg·m·s⁻¹ = N·s, the same units as Newtonian momentum. The argument v²/c² is (m·s⁻¹)²/(m·s⁻¹)² = dimensionless, as required inside the square root.

Limiting cases

Low speed v « c: γ ≈ 1 + ½v²/c², so p ≈ mv(1 + ½v²/c²) → mv: the Newtonian result, recovered to leading order.
Ultra-relativistic v → c: γ → ∞ and p → ∞; combined with E = γmc² this gives p ≈ E/c, the photon-like regime.
Massless limit m → 0 with v = c: the product γm is indeterminate (0×∞) and finite; p = E/c takes over, describing light.
Rest v = 0: γ = 1, p = 0, as required by isotropy.

Breaks when

Massless particles (m = 0). The form p = γmv gives 0×∞ and is useless; photons require p = E/c = ℏk directly, obtained from the energy–momentum invariant with m = 0.
Quantum / field regime. For a single quantum, momentum is an operator eigenvalue p = ℏk, not a function of a classical trajectory; the notion of a definite v fails and this classical expression no longer applies.
Curved spacetime / non-inertial frames. The derivation assumed global inertial frames linked by Lorentz boosts. In general relativity there is no single global boost symmetry, and momentum must be defined via the stress–energy tensor and parallel transport instead.
Composite or radiating systems. If the "rest mass" is not invariant — e.g. an object that emits radiation or an open system exchanging energy — m changes and the simple γmv bookkeeping must be replaced by the full four-momentum of all constituents.

Failure modes

"Relativistic mass" confusion: writing m_rel = γm and then using it in F = m_rela or E = m_relc² everywhere. It works for p = m_relv but fails for force, because F = dp/dt is not γma in general (the γ also varies).
Putting γ in the wrong place: using γ evaluated at the frame's boost speed rather than at the particle's own speed in the frame of interest.
Adding velocities Galileanly while keeping the relativistic γ — mixing the two frameworks and getting quantities that conserve in no frame.
Forgetting transverse-momentum invariance: transforming p_y as if it changed under an x-boost, which breaks the very step that fixes f.
Assuming p = mv conserves in relativity: it conserves in the collision's own CM frame but not after a boost, which is the whole point being missed.

Discussion

The logical structure here is worth dwelling on. Momentum conservation is not derived from relativity; it is imposed as a law we insist on keeping. What relativity supplies is a new rule for how velocities transform between frames. The tension between "conservation must hold in every frame" and "velocities add relativistically" has exactly one resolution for the momentum of a point particle, and that resolution is the factor γ. Change the transformation law back to Galilean and the tension vanishes — which is precisely why Newton never needed γ.

A cleaner and more modern route reaches the same place: define momentum as rest mass times the proper-time derivative of position, p^μ = m dx^μ/dτ = m U^μ (from proper-time-and-four-velocity). Because proper time τ is Lorentz-invariant and x^μ is a four-vector, p^μ is automatically a four-vector and its conservation is frame-independent by construction. The spatial part is m dx/dτ = m (dx/dt)(dt/dτ) = γmv, because dt/dτ = γ. The collision argument and the four-vector argument agree because they are enforcing the same requirement from two directions.

The time component of that same four-vector is p⁰ = γmc = E/c, so energy and momentum are two faces of one geometric object. Their invariant length gives E² − (pc)² = (mc²)², the relation every particle physicist uses. Thus fixing p = γmv is inseparable from mass–energy equivalence; you cannot accept one and reject the other.

At the deepest level, this connects to symmetry (a thread of this unit): Noether's theorem ties momentum conservation to invariance under spatial translations, and the requirement that this conservation law be Lorentz-covariant — i.e. that it hold identically in all boosted frames — is what upgrades the conserved quantity from a 3-vector to the spatial part of a covariant four-vector. The factor γ is the visible fingerprint of demanding that a translation symmetry and a boost symmetry coexist. Seen this way, γmv is less a formula to memorise than the unique intersection of two symmetry principles.

Common misconceptions. (i) "Mass increases with speed." Nothing about the particle's intrinsic rest mass m changes; only the momentum and energy grow, through γ. Modern usage keeps m invariant. (ii) "The γ is an experimental correction." No — it is logically forced once conservation and Lorentz kinematics are both required. (iii) "Momentum is capped because v is capped." False: v < c always, but p is unbounded because γ diverges.

Worked examples

Example 1 — Proton in the LHC. A proton (m = 1.673×10⁻²⁷ kg) moves at v = 0.999999991c. Find its momentum, and compare to the Newtonian estimate.

γ = 1/√(1 − v²/c²)

Compute the Lorentz factor first, symbolically, before inserting numbers. A

1 − v²/c² = (1 − v/c)(1 + v/c) ≈ (9.0×10⁻⁹)(2) = 1.8×10⁻⁸

Use the factorisation to avoid catastrophic cancellation; v/c = 0.999999991 so 1 − v/c = 9.0×10⁻⁹. B

γ = 1/√(1.8×10⁻⁸) ≈ 7.5×10³

Take the reciprocal square root. This proton carries a Lorentz factor of about 7500 (roughly 7 TeV total energy). A

p = γmv ≈ (7.5×10³)(1.673×10⁻²⁷ kg)(3.0×10⁸ m/s)

Insert numbers into the boxed result, with v ≈ c. A

p ≈ 3.8×10⁻¹⁵ kg·m/s

Reading. The Newtonian estimate mv ≈ 5.0×10⁻¹⁹ kg·m/s is about 7500 times too small — the whole momentum of an LHC proton is the γ factor. In convenient units this is pc ≈ 7 TeV.

Example 2 — Electron at 0.6c. An electron (m = 9.109×10⁻³¹ kg) moves at v = 0.60c. Find p and the fractional error of the Newtonian formula.

γ = 1/√(1 − 0.60²) = 1/√(1 − 0.36) = 1/√0.64

Insert v/c = 0.60 into the Lorentz factor. A

γ = 1/0.80 = 1.25

Evaluate the square root (√0.64 = 0.80). A moderate speed gives a modest 25% enhancement. A

v = 0.60(3.0×10⁸) = 1.8×10⁸ m/s

Convert the speed to SI. A

p = γmv = (1.25)(9.109×10⁻³¹)(1.8×10⁸)

Insert into the boxed result. A

p ≈ 2.05×10⁻²² kg·m/s

Reading. The Newtonian value mv = 1.64×10⁻²² kg·m/s underestimates by the factor γ = 1.25, a 20% error — already significant at 60% of light speed, showing the correction is not confined to extreme energies.

Problems

A particle of rest mass m has relativistic momentum equal in magnitude to mc. Find its speed v/c and its Lorentz factor.
Solution
Set γmv = mc ⇒ γ(v/c) = 1. With γ = 1/√(1 − β²) and β = v/c: β/√(1 − β²) = 1 ⇒ β² = 1 − β² ⇒ β² = ½ ⇒ v/c = 1/√2 ≈ 0.707. Then γ = 1/√(1 − ½) = √2 ≈ 1.414. (This is also the speed where pc = mc², i.e. kinetic-like scales cross.)
Show that v = pc²/E for any particle, using p = γmv and E = γmc². Then evaluate v/c for a particle with pc = 3 MeV and E = 5 MeV.
Solution
Divide the two: pc²/E = (γmv)c²/(γmc²) = v. Hence v = pc²/E, i.e. v/c = pc/E = 3/5 = 0.60. (Consistent with a rest energy mc² = √(E² − (pc)²) = √(25 − 9) = 4 MeV, so γ = 5/4 = 1.25 — matching β = 0.6.)
Two identical particles of rest mass m each move toward each other, each at speed 0.80c in the lab. Find the total momentum and total energy of the system in the lab frame, and hence the invariant mass of the pair.
Solution
By symmetry the momenta cancel: P_total = 0. Each has γ = 1/√(1 − 0.64) = 1/0.60 = 1.667, so each energy is γmc² = 1.667 mc²; total E_total = 3.33 mc². Invariant mass Mc² = √(E_total² − (P_totalc)²) = E_total = 3.33 mc², so M = 3.33m — larger than 2m because kinetic energy contributes to the system's invariant mass.
Starting from p = γmv, show that the relativistic force F = dp/dt for motion along a straight line equals γ³ma, where a = dv/dt.
Solution
With p = mv(1 − v²/c²)^−1/2, differentiate: dp/dt = m[(1 − β²)^−1/2 + v·(−½)(1 − β²)^−3/2(−2v/c²)] a = ma[(1 − β²)^−1/2 + (v²/c²)(1 − β²)^−3/2]. Factor (1 − β²)^−3/2: the bracket = (1 − β²)^−3/2[(1 − β²) + β²] = (1 − β²)^−3/2 = γ³. Hence F = γ³ma. This γ³ (longitudinal) is why "relativistic mass" fails for force.
A photon has no rest mass, yet carries momentum. A perfectly absorbing sail of area 1 m² sits at Earth's orbit where the solar flux is S = 1360 W/m². Using the photon relation p = E/c, find the radiation force on the sail. Why can't p = γmv be used here?
Solution
Power absorbed P = S·A = 1360 W. Momentum arrives at rate dp/dt = (dE/dt)/c = P/c = 1360/(3.0×10⁸) = 4.5×10⁻⁶ N. That momentum flux equals the force, so F ≈ 4.5 µN (double it for a perfectly reflecting sail). The formula p = γmv fails because a photon has m = 0 and v = c, giving the indeterminate 0×∞; only the energy–momentum relation p = E/c (the m → 0 limit of E² = (pc)² + (mc²)²) applies.